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So, anyway, I’m over here in my lab thinking about circuits and wonder about the inherent impedance of a 9 Volt battery.
To speak in greater detail, I’m wondering whether the impedance of the battery is low enough to consider it a useful path to signal ground in an audio circuit.
Otherwise I could add a bypass capacitor to bring signal ground up past the battery and viola.
Oh I know, let’s check Wikipedia.
Total posts : 45366The link gave a link for data on a 9 Volt Alkaline and here’s what it says:
LOAD (OHMS)
43K
620 PULSE.
Um, I used the term “inherent impedance” but…
I tend to think a “LOAD” is the circuit the battery is supplying voltage to, whereas the impedance of the battery itself would be a “source” impedance, which may equate with “inherent” impedance.
But wait. We have an alternate phantom power source and… I strongly suspect that a bypass capacitor would be recommended and since such a capacitor is probably advisable also with the battery, I think we’ve reached a decision.
That’s it for the day. No more thinking until tomorrow.
Total posts : 45366Is the opposition to the flow of alternating current. How can a DC device like a battery have an impedance?
Total posts : 45366In an audio circuit that is battery powered there is AC present in the form of the audio signal which is referenced to ground but the AC impedance of the battery will raise the AC component above signal ground by the internal AC resistance of the battery unless, as I’m here pre-supposing, a bypass electrolytic is used to bypass the battery and raise signal ground to the AC audio.
Everything has an internal impedance. Even our heads, which also function on DC with brain waves being AC.
DC is AC standing still.
Total posts : 45366My question pertains to 9 Volt batteries but this reading from the Instruction Manual for my Smart AA Battery Charger suggests that battery resistance is a thing:
“The charger will analyze the dynamic internal resistance by applying a load current and this current reading is referred to the voltage drop detected on the battery… etc.”
It seems to me both the DC and AC in a circuit need to complete a full circle (circuit) for a device to operate at optimum.
Total posts : 45366hmmmmm…. a new way to describe DC.
Mark
Total posts : 45366There is a classic symptom of the failure of the bypass capacitor across the battery in battery powered radios called motorboating. This is a low frequency oscillation which sounds like “putt…putt…putt..” hence the nickname.
What happens is when the bypass capacitor becomes weak the internal resistance of the battery allows the supply voltage to drop when the speaker is drawing current and this drop causes a decrease in the signal to the speaker which causes the supply voltage to rise which increases the signal to the speaker which causes the supply voltage to drop and so on. (was this as much fun to read as it was to type?) A good bypass capacitor will stabilize the supply voltage and prevent this oscillation. For this reason it is good practice to place a bypass capacitor across a battery supply.
As energy is extracted from a battery the internal resistance increases so this effect is most pronounced with a weak battery. A weak battery can have a normal terminal voltage under no load but still not be suitable for use. Batteries need to be tested under load for this reason. If, when under load, the terminal voltage is low then the battery needs replacement.
Neil
Total posts : 45366Thank you Neil for filling in with knowledge my technical books don’t mention.
Your opening sentence in paragraph 2 reads: “What happens is when the bypass capacitor becomes weak the internal resistance of the battery allows the supply voltage to drop…”
Did you mean to say “when the bypass capacitor becomes weak” or “when the battery becomes weak“?
If the former, what constitutes a weak bypass capacitor?
Total posts : 45366I was referring to the capicator becoming “weak” meaning the capacitance is reduced. This happens with age and temperature as the electrolyte dries out and the capacitance decreases and the ESR (internal resistance) increases so the cap is no longer effective in dampening the quick changes in voltage.
Neil
Total posts : 45366Thanks for the clarification.
I’m using 100 uF for the bypass but in other circuits in books a variety of values are used for the same purpose.
How is the value chosen?
Total posts : 45366There are equations which can be used to select a value based on the lowest freequency to suppress, the AC current expected, and the allowable voltage changes but for a one time prototype a rule of thumb I use is 1 uF for each mA of current supplied by the battery.
It would seem that bigger is better but the larger the capacitance generally the larger the ESR so it is a compromise.
Neil
Total posts : 45366From the “Give Credit Where Credit Is Due” department:
Neil, you are a better educator on these topics than this huge stack of books I have here which mainly put weight on the floorboards.
I think I can do all the things you suggested!
Feeling smarter isn’t that frequent a thing.
Total posts : 45366Neil advises: “For a one time prototype a rule of thumb I use is 1 uF for each mA of current supplied by the battery.”
My (2) electret microphone capsules are each rated at 1.0 mA, which easily works out to 2 uF for the bypass capacitor.
We have been talking about bypassing the battery, but there is also the matter of placing a bypass cap at the source of a FET (in some cases)… unless you say otherwise I will suppose that the same rule applies.
Total posts : 45366Ever notice some circuits will parallel a large electrolytic with a small value like .1 uf or less?
Electrolytics have an inductance which at higher frequencies raises the impedance. Think about how they’re made. Two long sheets of foil rolled into coils separated by insulation.
As such the reactance of the coil offsets the capacitive reactance.
Suppose an.electrolytic can have a resonant frequency like a coil?
Total posts : 45366I am not sure of your circuit but generally a source bypass cap is used in a common source amplifier where the signal is taken from the drain. In this case the cap functions to raise the AC gain of the amplifier and the value is based on the lower cutoff frequency desired. For a FET this is C = 1/(2*pi*R*f) where R is the source resistance. For a bipolar transistor it is a bit more complicated since the resistances seen by the base appear at the emitter and have to be accounted.
If your signal is taken from the source (a source follower) then no bypass cap should be used.
Neil
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